Question: Rewrite the equation by completing the square. $2x^{2}-3x-5 = 0$ $(x + $
Solution: $\begin{aligned} 2x^2-3x-5&=0 \\\\ 2x^2-3x&=5 \\\\ x^2-\dfrac32x&=\dfrac52 \end{aligned}$ Now we want to complete $x^2-\dfrac32x$ into a perfect square. To do that, we should add $\left(\dfrac{{-\frac32}}{2}\right)^2={\dfrac{9}{16}}$ to it: $x^2{-\dfrac32}x+{\dfrac{9}{16}}=\left(x-\dfrac34\right)^2$ $\begin{aligned} x^2-\dfrac32x&=\dfrac52 \\\\ x^2-\dfrac32x+{\dfrac{9}{16}}&=\dfrac52+{\dfrac{9}{16}} \\\\ \left(x-\dfrac34\right)^2&=\dfrac{49}{16} \end{aligned}$ In conclusion, the equation after completing the square is written as $\left(x-\dfrac34\right)^2=\dfrac{49}{16}$